\(\int \frac {(2-b x)^{3/2}}{\sqrt {x}} \, dx\) [542]

Optimal result

Integrand size = 16, antiderivative size = 63 \[ \int \frac {(2-b x)^{3/2}}{\sqrt {x}} \, dx=\frac {3}{2} \sqrt {x} \sqrt {2-b x}+\frac {1}{2} \sqrt {x} (2-b x)^{3/2}+\frac {3 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \]

[Out]

3*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(1/2)+1/2*(-b*x+2)^(3/2)*x^(1/2)+3/2*x^(1/2)*(-b*x+2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {52, 56, 222} \[ \int \frac {(2-b x)^{3/2}}{\sqrt {x}} \, dx=\frac {3 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}}+\frac {1}{2} \sqrt {x} (2-b x)^{3/2}+\frac {3}{2} \sqrt {x} \sqrt {2-b x} \]

[In]

Int[(2 - b*x)^(3/2)/Sqrt[x],x]

[Out]

(3*Sqrt[x]*Sqrt[2 - b*x])/2 + (Sqrt[x]*(2 - b*x)^(3/2))/2 + (3*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/Sqrt[b]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 56

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \sqrt {x} (2-b x)^{3/2}+\frac {3}{2} \int \frac {\sqrt {2-b x}}{\sqrt {x}} \, dx \\ & = \frac {3}{2} \sqrt {x} \sqrt {2-b x}+\frac {1}{2} \sqrt {x} (2-b x)^{3/2}+\frac {3}{2} \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx \\ & = \frac {3}{2} \sqrt {x} \sqrt {2-b x}+\frac {1}{2} \sqrt {x} (2-b x)^{3/2}+3 \text {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {3}{2} \sqrt {x} \sqrt {2-b x}+\frac {1}{2} \sqrt {x} (2-b x)^{3/2}+\frac {3 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02 \[ \int \frac {(2-b x)^{3/2}}{\sqrt {x}} \, dx=-\frac {1}{2} \sqrt {x} \sqrt {2-b x} (-5+b x)-\frac {6 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{\sqrt {b}} \]

[In]

Integrate[(2 - b*x)^(3/2)/Sqrt[x],x]

[Out]

-1/2*(Sqrt[x]*Sqrt[2 - b*x]*(-5 + b*x)) - (6*ArcTan[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 - b*x])])/Sqrt[b]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.10

method	result	size
meijerg	\(-\frac {3 \sqrt {-b}\, \left (\frac {4 \sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {-b}\, \left (-\frac {b x}{8}+\frac {5}{8}\right ) \sqrt {-\frac {b x}{2}+1}}{3}+\frac {\sqrt {\pi }\, \sqrt {-b}\, \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{\sqrt {b}}\right )}{\sqrt {\pi }\, b}\)	\(69\)
default	\(\frac {\left (-b x +2\right )^{\frac {3}{2}} \sqrt {x}}{2}+\frac {3 \sqrt {x}\, \sqrt {-b x +2}}{2}+\frac {3 \sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right )}{2 \sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}\)	\(78\)
risch	\(\frac {\left (b x -5\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{2 \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {3 \sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right )}{2 \sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}\)	\(95\)

[In]

int((-b*x+2)^(3/2)/x^(1/2),x,method=_RETURNVERBOSE)

[Out]

-3*(-b)^(1/2)/Pi^(1/2)/b*(4/3*Pi^(1/2)*x^(1/2)*2^(1/2)*(-b)^(1/2)*(-1/8*b*x+5/8)*(-1/2*b*x+1)^(1/2)+Pi^(1/2)*(
-b)^(1/2)/b^(1/2)*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.70 \[ \int \frac {(2-b x)^{3/2}}{\sqrt {x}} \, dx=\left [-\frac {{\left (b^{2} x - 5 \, b\right )} \sqrt {-b x + 2} \sqrt {x} + 3 \, \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{2 \, b}, -\frac {{\left (b^{2} x - 5 \, b\right )} \sqrt {-b x + 2} \sqrt {x} + 6 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{2 \, b}\right ] \]

[In]

integrate((-b*x+2)^(3/2)/x^(1/2),x, algorithm="fricas")

[Out]

[-1/2*((b^2*x - 5*b)*sqrt(-b*x + 2)*sqrt(x) + 3*sqrt(-b)*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1))/b, -
1/2*((b^2*x - 5*b)*sqrt(-b*x + 2)*sqrt(x) + 6*sqrt(b)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))))/b]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.03 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.62 \[ \int \frac {(2-b x)^{3/2}}{\sqrt {x}} \, dx=\begin {cases} - \frac {i b^{2} x^{\frac {5}{2}}}{2 \sqrt {b x - 2}} + \frac {7 i b x^{\frac {3}{2}}}{2 \sqrt {b x - 2}} - \frac {5 i \sqrt {x}}{\sqrt {b x - 2}} - \frac {3 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{\sqrt {b}} & \text {for}\: \left |{b x}\right | > 2 \\\frac {b^{2} x^{\frac {5}{2}}}{2 \sqrt {- b x + 2}} - \frac {7 b x^{\frac {3}{2}}}{2 \sqrt {- b x + 2}} + \frac {5 \sqrt {x}}{\sqrt {- b x + 2}} + \frac {3 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{\sqrt {b}} & \text {otherwise} \end {cases} \]

[In]

integrate((-b*x+2)**(3/2)/x**(1/2),x)

[Out]

Piecewise((-I*b**2*x**(5/2)/(2*sqrt(b*x - 2)) + 7*I*b*x**(3/2)/(2*sqrt(b*x - 2)) - 5*I*sqrt(x)/sqrt(b*x - 2) -
3*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/sqrt(b), Abs(b*x) > 2), (b**2*x**(5/2)/(2*sqrt(-b*x + 2)) - 7*b*x**(3/2)
/(2*sqrt(-b*x + 2)) + 5*sqrt(x)/sqrt(-b*x + 2) + 3*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/sqrt(b), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.25 \[ \int \frac {(2-b x)^{3/2}}{\sqrt {x}} \, dx=-\frac {3 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{\sqrt {b}} + \frac {\frac {3 \, \sqrt {-b x + 2} b}{\sqrt {x}} + \frac {5 \, {\left (-b x + 2\right )}^{\frac {3}{2}}}{x^{\frac {3}{2}}}}{b^{2} - \frac {2 \, {\left (b x - 2\right )} b}{x} + \frac {{\left (b x - 2\right )}^{2}}{x^{2}}} \]

[In]

integrate((-b*x+2)^(3/2)/x^(1/2),x, algorithm="maxima")

[Out]

-3*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x)))/sqrt(b) + (3*sqrt(-b*x + 2)*b/sqrt(x) + 5*(-b*x + 2)^(3/2)/x^(3/2)
)/(b^2 - 2*(b*x - 2)*b/x + (b*x - 2)^2/x^2)

Giac [A] (verification not implemented)

none

Time = 6.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.32 \[ \int \frac {(2-b x)^{3/2}}{\sqrt {x}} \, dx=-\frac {{\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} {\left (\frac {b x - 2}{b} - \frac {3}{b}\right )} - \frac {6 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b}}\right )} b}{2 \, {\left | b \right |}} \]

[In]

integrate((-b*x+2)^(3/2)/x^(1/2),x, algorithm="giac")

[Out]

-1/2*(sqrt((b*x - 2)*b + 2*b)*sqrt(-b*x + 2)*((b*x - 2)/b - 3/b) - 6*log(abs(-sqrt(-b*x + 2)*sqrt(-b) + sqrt((
b*x - 2)*b + 2*b)))/sqrt(-b))*b/abs(b)

Mupad [F(-1)]

Timed out. \[ \int \frac {(2-b x)^{3/2}}{\sqrt {x}} \, dx=\int \frac {{\left (2-b\,x\right )}^{3/2}}{\sqrt {x}} \,d x \]

[In]

int((2 - b*x)^(3/2)/x^(1/2),x)

[Out]

int((2 - b*x)^(3/2)/x^(1/2), x)